问题 选择题

已知两圆半径r1、r2分别是方程x2﹣7x+10=0的两根,两圆的圆心距为7,则两圆的位置关系是(  )

A.相交

B.内切

C.外切

D.外离

答案

C.

题目分析:首先解方程x2-7x+10=0,求得两圆半径r1、r2的值,又由两圆的圆心距为7,根据两圆位置关系与圆心距d,两圆半径R,r的数量关系间的联系即可得出两圆位置关系.

∵x2-7x+10=0,

∴(x-2)(x-5)=0,

∴x1=2,x2=5,

即两圆半径r1、r2分别是2,5,

∵2+5=7,故两圆的位置关系是外切.

故选:C.

考点:1. 圆与圆的位置关系;2.解一元二次方程-因式分解法.

阅读理解

阅读理解。

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单项选择题